Regex to get a specific function code from php file

Question

I'm editing a PHP file to replace a function's code by another code provided. My problem is that the function's code may change. So, I don't know exactly how I should do the regex to find it.

Here's an example:

function lol()
{
    echo('omg');
    $num=0;
    while($num<100)
    {
        echo('ok');
    }
}
function teste()
{
    echo('ola');
    while($num<100)
    {
        echo('ok');
    }
}
function test()
{
    echo('oi');
    while($num<100)
    {
        echo('ok');
    }
}

What I need is to get the code of function teste(), which would be:

echo('ola');
while($num<100)
{
    echo('ok');
}

BUT: I do not know how many whiles, foreaches or how many brackets are inside the functions, neither I don't know it's order. The example was just an example.

How would I be able to get the function's code? Thank you in advance.

What is your final goal with this? There may be a better way of doing what you want — Pekka
– Pekka, Commented Jun 25, 2013 at 8:05
do you need EVERYTHING is inside the function teste() or just the first line? — STT LCU
– STT LCU, Commented Jun 25, 2013 at 8:05
Do you want the full content of your function of just the echo? — Édouard Lopez
– Édouard Lopez, Commented Jun 25, 2013 at 8:10

Édouard Lopez · Accepted Answer · 2013-06-25 10:47:41Z

2

Disclaimer

As other users stated, this is probably not a good approach. And if you decided to use it anyway, double-check your result as regex can be treacherous.

Ugly solution

You can you something like to match even if the function is the last one (@Nippey):

(?:function teste\(\))[^{]*{(.*?)}[^}]*?(?:function|\Z)

Note: I'm using (?:xyz) which is for non-capturing parentheses group. Check if your regex engine support it.

Output

    echo('ola');
    while($num<100)
    {
        echo('ok');
    }

edited Jun 25, 2013 at 10:47

answered Jun 25, 2013 at 8:27

Édouard Lopez

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5 Comments

STT LCU Over a year ago

I know you explicitly stated that this works if teste is not the last function... but what happens if it is?

Nippey Over a year ago

You could use this modification, so you don't have to remove the brackets: (?:function teste\(\))[^{]*{(.*?)}[^}]*?(?:function)

Nippey Over a year ago

@STTLCU: Use this: (?:function teste\(\))[^{]*{(.*?)}[^}]*?(?:function|\Z)

STT LCU Over a year ago

@EdouardLopez i'm not OP but i guess that if you used nippey's suggestion now the output is without external braces

user2271353 Over a year ago

I used the first option, where function is not the last function, as I know it isn't and won't be in the platform which with I'm working for. However, this last one is better, so thank you both of you :)

HamZa · Accepted Answer · 2013-06-25 09:08:17Z

1

Using recursive pattern (?R) you can match nested brackets:

$result = preg_replace_callback('#(.*?)\{((?:[^{}]|(?R))*)\}|.*#si', function($m){
    if(isset($m[1], $m[2]) && preg_match('#function\s+teste\b#i', $m[1])){
        return $m[2];
    }
}, $code); // Anonymous function is used here which means PHP 5.3 is required

echo $result;

Online demo

answered Jun 25, 2013 at 9:08

HamZa

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